{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Euler Problem\n", "\n", "2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.\n", "\n", "What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?\n", "\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "My easiest guess is to multiply all prime numbers till the number." ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": true }, "outputs": [], "source": [ "def get_primes_smaller(number):\n", " primes = []\n", " prospects = [n for n in range(2, number)]\n", " while prospects:\n", " p = prospects[0]\n", " prospects = [x for x in prospects if x % p != 0]\n", " primes.append(p)\n", " return primes" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "210\n" ] } ], "source": [ "from operator import mul\n", "from functools import reduce\n", "\n", "def get_number_which_is_divisible_by_all_numbers_from_one_to(n):\n", " ps= get_primes_smaller(n + 1)\n", " return reduce(mul, ps, 1)\n", "\n", "print(get_number_which_is_divisible_by_all_numbers_from_one_to(10))" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "That obviously didn't work. The reason is that the same prime can occur multiple times in the factorization of a divisor. For example $2^{3} = 8$. We can always brute force of course. We do a smart brute force and only check multiples from the product of primes because this factor must be part of the solution." ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "232792560\n" ] } ], "source": [ "def is_divisible_by_numbers_smaller_or_equal(number, maximum_number):\n", " for n in range(2, maximum_number + 1):\n", " if number % n != 0:\n", " return False\n", " return True\n", "\n", "def get_number_which_is_divisible_by_all_numbers_from_one_to(n):\n", " ps = get_primes_smaller(n + 1)\n", " factor = reduce(mul, ps, 1)\n", " multiples_of_factor = factor\n", " while True:\n", " if is_divisible_by_numbers_smaller_or_equal(multiples_of_factor, n):\n", " return multiples_of_factor\n", " multiples_of_factor += factor\n", "\n", "assert(get_number_which_is_divisible_by_all_numbers_from_one_to(10) == 2520)\n", "print(get_number_which_is_divisible_by_all_numbers_from_one_to(20))" ] } ], "metadata": { "completion_date": "Wed, 20 Aug 2014, 14:32", "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.5.4" }, "tags": [ "reduce", "divisible" ] }, "nbformat": 4, "nbformat_minor": 0 }